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Zina [86]
1 year ago
14

A student uses a dye to change the colour of his tee-shirt. he fills a metal bucket with water. When the bucket is heated, what

happens to the movement of the particles in the metal?
Chemistry
1 answer:
Tanya [424]1 year ago
8 0

Answer:

The particles begin to vibrate faster and more.

Explanation:

Adding heat to matter increases the energy, thus creating more movement. Eventually, the bucket will melt, turning to a liquid. While it is a sold, it still has particle movement, just not enough to break volume or shape.

You might be interested in
The values used in the scale of pH and pOH are derived from a system designed by ______. Gordonsen Sorenson Curie Dalton
yaroslaw [1]

Sorenson

Explanation:

The values used in the scale of pH and pOH are derived from a system designed by Sorenson. Søren Peter Lauritz Sørensen, a Danish chemist introduced the system of pH and pOH for describing the alkalinity and acidity of substances.

  • The pH and pOH scale is logarithmic scale that ranks the acidity and bascity of compounds.
  • pH is the negative logarithm of the concentration of  hydrogen/hydroxonium ions in solution i.e

              pH = -log₁₀{H⁺]

  • pOH is the negative log of the concentration of the hydroxyl ions in a solution i.e

             pOH = -log₁₀{OH⁻]

Learn more:

calculating pH: brainly.com/question/12985875

pH scale: brainly.com/question/11063271

#learnwithBrainly

3 0
2 years ago
32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?
kotegsom [21]
Under standard temperature and pressure conditions, it is known that 1 mole of a gas occupies 22.4 liters.

From the periodic table:
molar mass of oxygen = 16 gm
molar mass of hydrogen = 1 gm
Thus, the molar mass of water vapor = 2(1) + 16 = 18 gm

18 gm of water occupies 22.4 liters, therefore:
volume occupied by 32.7 gm = (32.7 x 22.4) / 18 = 40.6933 liters

5 0
1 year ago
3. Which of these statements is an accurate description of the ionization energies of elements in the periodic table?
dalvyx [7]

Answer:

option A is correct

Explanation:

7 0
2 years ago
If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500
ratelena [41]

<u>Answer:</u> The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

                         SO_2+NO_2\rightleftharpoons SO_3+NO

<u>Initial:</u>             0.500  0.500      0.005   0.005

<u>At eqllm:</u>      0.500-x  0.500-x   0.005+x  0.005+x

The expression of equilibrium constant for the above reaction follows:

K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}

We are given:

K_c=2.50

Putting values in above equation, we get:

2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = (0.500-x)=(0.500-0.304)=0.196M

Equilibrium concentration of nitrogen dioxide = (0.500-x)=(0.500-0.304)=0.196M

Equilibrium concentration of sulfur trioxide = (0.00500+x)=(0.00500+0.304)=0.309M

Equilibrium concentration of nitrogen monoxide = (0.00500+x)=(0.00500+0.304)=0.309M

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

4 0
2 years ago
Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:
Katarina [22]

Answer and Explanation:

The 3 reactions represented are

C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O

C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O

C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O

ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol

ΔH°f = ΔH°(products) - ΔH°(reactants)

For reaction 1,

ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol

For reaction 2,

ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol

For reaction 3,

ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol

b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.

Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol

Number of moles = 254/296 = 0.858 mole

heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol

c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole

5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole

The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole

q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ

QED!!!

8 0
1 year ago
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